问题描述
这个问题是有关：
DSP ：音频处理：squart或登录利用FFT
在我损失了大约正确的算法来选择。
in which I was lost about the right algorithm to choose.
现在，
我要得到我的信号的所有频率，我从一个音频文件得到的。
I want to get all the frequencies of my signal, that I get from an audio file.
我使用numpy的，和scikits.audiolab。我做了阅读在DSP上受到了不少，去dspguru.com以及阅读过网文件和漂亮的博客。
I use numpy, and scikits.audiolab. I made a lot of reading on the dsp subject, went to dspguru.com as well, read papers and nice blogs over the net.
在code我用这一条是：
The code I use is this one :
import numpy as np
from scikits.audiolab import Sndfile
f = Sndfile('first.ogg', 'r')
# Sndfile instances can be queried for the audio file meta-data
fs = f.samplerate
nc = f.channels
enc = f.encoding
print(fs,nc,enc)
# Reading is straightfoward
data = f.read_frames(10)
print(data)
print(np.fft.rfft(data))
我是新来的DSP。
I am new to DSP.
我想能够的信号的所有频率分离，以比较不同的信号。
我用numpy.fft.rfft声音的阵列上;但现在，仅此操作是不够的。那么，什么是获得频率的所有大小的最佳解决方案是否正确？
I would like to be able to separate all the frequencies of a signal to compare different signals.I use numpy.fft.rfft on the array of sound; But now, this operation alone is not enough. So, what is the best solution to get all the magnitudes of frequencies correctly ?
我看到相乘所得到的值获得复数关闭，整个转化为实数。
I saw that multiplying the resulting values get the complex numbers off and transform the whole as a real number.
现在是什么吗？是不是这样？
Now what please ? Is that it ?
如果你需要我澄清什么，只是问。
if you need me to clarify anything, just ask.
非常感谢！
推荐答案
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您说我想我的信号的所有频率，我从一个音频文件得到的。但你真正想要的是频率的幅度。
You say "I want to get all the frequencies of my signal, that I get from an audio file." but what you really want is the magnitude of the frequencies.
在您的code，它看起来像（我不知道蟒蛇）你只读了前10个样本。假设你的文件是单声道的，这很好，但你可能想看看更大的一组样品，说1024个样本。一旦你这样做，当然，你要重复上下一组N个样本。您可能会或可能不想重叠样本集，您可能希望应用窗口的功能，但你所做的是一个良好的开端。
In your code, it looks like (I don't know python) you only read the first 10 samples. Assuming your file is mono, that's fine, but you probably want to look at a larger set of samples, say 1024 samples. Once you do that, of course, you'll want to repeat on the next set of N samples. You may or may not want to overlap the sets of samples, and you may want to apply a window function, but what you've done here is a good start.
什么贪睡说的是真的。 FFT的输出是复杂的。为了找到一个给定频率的大小，你需要找到长度或复数，这简直是开方的绝对值（R ^ 2 + I ^ 2）。
What sleepyhead says is true. The output of the fft is complex. To find the magnitude of a given frequency, you need to find the length or absolute value of the complex number, which is simply sqrt( r^2 + i^2 ).
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